Question: $g(x)=\dfrac{1}{1-x}$ We know that $g(x)=1+x+x^2+x^3+...$ for $x\in(-1,1)$. Using this fact, find the power series for $f(x)=\dfrac{1}{(1-x)^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1+x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3!}+...$ (Choice B) B $(1-x)-\frac{{{(1-x)}^{2}}}{2}+\frac{{{(1-x)}^{3}}}{3!}+...$ (Choice C) C $x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3!}+...$ (Choice D) D $1+2x+3{{x}^{2}}+4{{x}^{3}}+...$ (Choice E) E $1+{{x}^{2}}+...$
Solution: First, notice that the derivative of $ ~g(x)=\frac{1}{1-x}~~$ is $~~\frac{1}{{{(1-x)}^{2}}}\,$. This is precisely the function $~f(x)~$ that we are looking for. From our knowledge of geometric series, we know the following: $\frac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}+...$ Taking the derivative of both sides gives us what you see below. $\frac{1}{{{(1-x)}^{2}}}=1+2x+3{{x}^{2}}+4{{x}^{3}}+...$